Bash copy variable containing array as reference

Question

With the following code I can copy an array content to another one so that when altering one array the other is untouched.

#!/bin/bash
declare -a ARRAY_A=(a b c d e)
ARRAY_B=("${ARRAY_A[@]}")

echo "Before removal:"

printf "%s " ${ARRAY_A[@]}
echo
printf "%s " ${ARRAY_B[@]}
echo

echo
echo "After removal:"
unset ARRAY_A[0]

printf "%s " ${ARRAY_A[@]}
echo
printf "%s " ${ARRAY_B[@]}
echo

Prints:

Before removal:
a b c d e 
a b c d e 

After removal:
b c d e 
a b c d e

Is it possible to copy the reference to that array instead, so that when altering one array the "other" (which is the same then) appears to be changed as well (like below)?

Before removal:
a b c d e 
a b c d e 

After removal:
b c d e 
b c d e

KamilCuk · Accepted Answer · 2020-09-21 07:11:36Z

5

Yes? Just use a bash name reference.

#!/bin/bash
declare -a ARRAY_A=(a b c d e)
declare -n ARRAY_B=ARRAY_A

echo "Before removal:"

printf "%s " "${ARRAY_A[@]}"
echo
printf "%s " "${ARRAY_B[@]}"
echo

echo
echo "After removal:"
unset ARRAY_A[0]

printf "%s " "${ARRAY_A[@]}"
echo
printf "%s " "${ARRAY_B[@]}"
echo

Note: prefer to use lowercase variables in your scripts. Upper case variables are by convention meant to be exported variables, like COLUMN, LINES, PWD, UID, IFS, etc.

answered Sep 21, 2020 at 7:11

KamilCuk

146k8 gold badges84 silver badges154 bronze badges

Sign up to request clarification or add additional context in comments.

Collectives™ on Stack Overflow

Bash copy variable containing array as reference

1 Answer 1

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

Comments

Your Answer

Sign up or log in

Post as a guest

Related